Note that √(4 - t²) is defined only as long as 4 - t² ≥ 0, or -2 ≤ t ≤ 2. Then the real integral exists only if -2 ≤ x ≤ 2. (Otherwise we deal with complex numbers.)
If x = 2, then the integral corresponds to the area of a quarter-circle with radius 2. This means that the integral has a maximum value of 1/4 • π • 2² = π.
On the opposite end, if x = -2, then the integral has the same value, but the integral from 0 to -2 is equal to the negative integral from -2 to 0. So the minimum value is -π.
For all x in between, we observe that the integrand is continuous over the rest of its domain, so F(x) is continuous.
Then the range of F(x) is the interval [-π, π].
Answer:
1.11
Step-by-step explanation:
1.03+0.08=1.11
Answer:
square root(-1) = i
Step-by-step explanation:
Answer:
1
Step-by-step explanation:
:)..................
- for line B so the set of equations to have no solutions bc. x+y=2 and so y=2-x
so from x+y=4 will result y=4-x
so this is the right equation for line B so the set of equations has no solutions
